In part of an Artificial Neural Network matlab code, I want to find nearest points of two convex polygons.
I saw
dsearchn(X,T,XI)
command's description here, but that finds closest points between two sets of points, and polygons (like convexs) have infinites points.
So can you suggest any way/idea?
Notice: I'm using MATLAB 2014a. I have the coordinate of each convex's vertex point.
If you are not happy with what is provided by dsearchn, then, If I were you, I would do one of two following:
Find Nearest Neighbours on the vertices (for example which vertex of
polygon A is the NN of a given vertex of polygon B).
Pick a random point inside polygon A (you may want to compute the
convex hull of A, but you may skip that and take into account only
the vertices you know already). That random point is the query. Find
an NN of that point from the vertices of polygon B.
You may want to ask in Software recommendations for more.
Edit:
Another approach is this:
Create a representative dataset of polygon A. Set the size of the dataset yourself and fill it with samples of points that lie inside the polygon. Choose them uniformly randomly inside the polygon.
Then take a point of polygon B (either a vertex or a random point inside polygon B) and that's the query point, for which you will seek Nearest Neighbour(s) inside the representative dataset of polygon A.
Of course that's just an approximation, but I can't think of something else now.
Notice that you can of course do the same for polygon B.
With This File in File Exchange, I've find the solution.
With a little code modification, I have drawn a perpendicular bisector which I wanted. Of course, this way is time consuming.
Related
Suppose I want to uniformly sample points inside a convex polygon.
One of the most common approaches described here and on the internet in general consists in triangulation of the polygon and generate uniformly random points inside each triangles using different schemes.
The one I find most practical is to generate exponential distributions from uniform ones taking -log(U) for instance and normalizing the sum to one.
Within Matlab, we would have this code to sample uniformly inside a triangle:
vertex=[0 0;1 0;0.5 0.5]; %vertex coordinates in the 2D plane
mix_coeff=rand(10000,size(vertex,1)); %uniform generation of random coefficients
x=-log(x); %make the uniform distribution exponential
x=bsxfun(#rdivide,x,sum(x,2)); %normalize such that sum is equal to one
unif_samples=x*vertex; %calculate the 2D coordinates of each sample inside the triangle
And this works just fine:
However, using the exact same scheme for anything other than a triangle just fails. For instance for a quadrilateral, we get the following result:
Clearly, sampling is not uniform anymore and the more vertices you add, the more difficult it is to "reach" the corners.
If I triangulate the polygon first then uniform sampling in each triangle is easy and obviously gets the job done.
But why? Why is it necessary to triangulate first?
Which specific property have triangle (and simplexes in general since this behaviour seems to extend to n-dimensional constructions) that makes it work for them and not for the other polygons?
I would be grateful if someone could give me an intuitive explanation of the phenomena or just point to some reference that could help me understand what is going on.
I should point out that it's not strictly necessary to triangulate a polygon in order to sample uniformly from it. Another way to sample a shape is rejection sampling and proceeds as follows.
Determine a bounding box that covers the entire shape. For a polygon, this is as simple as finding the highest and lowest x and y coordinates of the polygon.
Choose a point uniformly at random in the bounding box.
If the point lies inside the shape, return that point. (For a polygon, algorithms that determine this are collectively called point-in-polygon predicates.) Otherwise, go to step 2.
However, there are two things that affect the running time of this algorithm:
The time complexity depends greatly on the shape in question. In general, the acceptance rate of this algorithm is the volume of the shape divided by the volume of the bounding box. (In particular, the acceptance rate is typically very low for high-dimensional shapes, in part because of the curse of dimensionality: typical shapes cover a much smaller volume than their bounding boxes.)
Also, the algorithm's efficiency depends on how fast it is to determine whether a point lies in the shape in question. Because of this, it's often the case that complex shapes are made up of simpler shapes, such as triangles, circles, and rectangles, for which it's easy to determine whether a point lies in the complex shape or to determine that shape's bounding box.
Note that rejection sampling can be applied, in principle, to sample any shape of any dimension, not just convex 2-dimensional polygons. It thus works for circles, ellipses, and curved shapes, among others.
And indeed, a polygon could, in principle, be decomposed into a myriad of shapes other than triangles, one of those shapes sampled in proportion to its area, and a point in that shape sampled at random via rejection sampling.
Now, to explain a little about the phenomenon you give in your second image:
What you have there is not a 4-sided (2-dimensional) polygon, but rather a 3-dimensional simplex (namely a tetrahedron) that was projected to 2-dimensional space. (See also the previous answer.) This projection explains why points inside the "polygon" appear denser in the interior than in the corners. You can see why if you picture the "polygon" as a tetrahedron with its four corners at different depths. With higher dimensions of simplex, this phenomenon becomes more and more acute, again due partly to the curse of dimensionality.
Well, there are less expensive methods to sample uniform in the triangle. You're sampling Dirichlet distribution in the simplex d+1 and taking projection, computing exponents and such. I would refer you to the code sample and paper reference here, only square roots, a lot simpler algorithm.
Concerning uniform sampling in complex areas (quadrilateral in your case) general approach is quite simple:
Triangulate. You'll get two triangles with vertices (a,b,c)0 and (a,b,c)1
Compute triangle areas A0 and A1 using, f.e. Heron's formula
First step, randomly select one of the triangles based on area.
if (random() < A0/(A0+A1)) select triangle 0 else select triangle 1. random() shall return float in the range [0...1]
Sample point in selected triangle using method mentioned above.
This approach could be easily extended to sample for any complex area with uniform density: N triangles, Categorical distribution sampling with probabilities proportional to areas will get you selected triangle, then sample point in the triangle.
UPDATE
We have to triangulate because we know good (fast, reliable, only 2 RNG calls, ...) algorithm to sample with uniform density in triangle. Then we could build on it, good software is all about reusability, and pick one triangle (at the cost of another rng call) and then back to sample from it, total three RNG calls to get uniform density sampling from ANY area, convex and concave alike. Pretty universal method, I would say. And triangulation is a solved problem, and
basically you do it once (triangulate and build weights array Ai/Atotal) and sample till infinity.
Another part of the answer is that we (me, to be precise, but I've worked with random sampling ~20years) don't know good algorithm to sample precisely with uniform density from arbitrary convex more-than-three-vertices closed polygon. You proposed some algorithm based on hunch and it didn't work out. And it shouldn't work, because what you use is Dirichlet distribution in d+1 simplex and project it back to d hyperplane. It is not extendable even to quadrilateral, not talking to some arbitrary convex polygon. And I would state conjecture, that even such algorithm exist, n-vertices polygon would require n-1 calls to RNG, which means there is no triangulation setup, but each call to get a point would be rather expensive.
Few words on complexity of the sampling. Assuming you did triangulation, then with 3 calls to RNG you'll get one point sampled uniformly inside your polygon.
But complexity of sampling wrt number of triangles N would be O(log(N)) at best. YOu basically would do binary search over partial sums of Ai/Atotal.
You could do a bit better, there is O(1) (constant time) sampling using Alias sampling of the triangle. The cost would be a bit more setup time, but it could be fused with triangulation. Also, it would require one more RNG calls. So for four RNG calls you would have constant point sampling time independent of complexity of your polygon, works for any shape
I'm working on a science project. I have a list of xyz-coordinates of voronoi diagram vertices, and a list of points that create my protein's convex-hull (from triangulation). Now some of the vertices lie quite far from the hull and I'd like to filter them out. How can I do it in c++ ? For now it doesn't have to be super optimised, I'm only focused on removing those points.
visualization
I was also thinking to check if a line from voronoi vertex(red crosses) to center of protein(pink sphere) is intersecting with hull face at any point, but I'm not sure how to achive that.
I've read that you can check if a point is inside a polygon by counting the times an infinite line from the point is crossing the hull, but that was for two dimensions. Can similar approach be implemented to suit my needs ?
https://www.geeksforgeeks.org/how-to-check-if-a-given-point-lies-inside-a-polygon/
Let's start with the two-dimensional case. You can find the solution to that on CS.SX:
Determine whether a point lies in a convex hull of points in O(logn)
The idea is that each two consecutive points on the convex hull define a triangular slice of the plane (to some internal point of the hull); you can find the slice in which your point is located, and check whether it's closer to the internal point than the line segment bounding the slide.
For the three-dimensional case, I'm guessing you should be able to do something similar, but the search for the 3 points forming the relevant triangle might be a little tricky. In particular, it would also depend on how the convex hull is represented - as in the 2-d case the convex hull is just a sequence of consecutive points on a cycle.
I know this doesn't quite solve your problem but it's the best I've got given what you've written...
Context: I want to create an interactive heatmap with areas separated by a ZIP code. I've found no way of displaying it directly (i.e. using Google Maps or OSM), so I want to create curves or lines that are separating those areas, and visualize it in maps.
I have a set of points, represented by their coordinates and their according class (ZIP code). I want to get a curve separating them. The problem is that these points are not linearly separable.
I tried to use softmax regression, but that doesn't work well with non-linearly separable classes. The only methods I know which are able to separate non-linearly are nearest neighbors and neural networks. But such classifiers only classify, they don't tell me the borders between classes.
Is there a way to get the borders somehow?
If you have a dense cloud of known points within each Zip code with coordinates [latitude. longitude, zip code], using machine learning to find the boundary enclosing those points sounds like overkill.
You could probably get a good approximation of the boundary by using computational geometry, e.g finding the 2D convex hull of each Zip code's set of points using the Matlab convhull function
K = convhull(X,Y)
The result K would be a vector of points enclosing the input X, Y vector of points, that could be used to draw a polygon.
The only complication would be what coordinate system to work in, you might need to do a bit of work going between (lat, lon) and map (x,y) coordinates. If you do not have the Matlab Mapping Toolbox, you could look at the third party library M_Map M_Map home page, which offers some of the same functionality.
Edit: If the cloud of points for Zip codes has a bounding region that is non convex, you may need a more general computational geometry technique to find a better approximation to the bounding region. Performing a Voronoi tesselation of the region, as suggested in the comments, is one such possibility.
I have to create a function in MATLAB that performs the following task:
Input:
p polygon in the form
p = [x1,y1; x2,y2; x3,y3; x4,y4...]
s struct with the segment from A to B
s = struct('A',[x,y],'B'[u,w])
Return:
1) An integer indicating how many intersections there are between the segment and the polygon (e.g., 0,1,2)
2) A new segment from A to B, where A is the first intersection or the initial point of the input segment and B the second point of the intersection or the last point of the segment input.
I have an idea on how to do it by using the function inpolygon. I have been reading how to use this function, and know that to use that, I should provide a query point and the coordinates of the polygon vertices. It will return 1 or 0 depending on whether it is inside or not.
My question is, how can I get the query point of the segment that is placed exactly in the boundary (in the case that the segment intersects with it)?
If you have the Mapping Toolbox installed, you could use polyxpoly. As this is a rather basic problem, there are quite a few free MATLAB-codes out there on the File Exchange. Here is what I found for the search term 'polygon intersect':
2D Polygon edges intersection by Bruno Luong
Find the intersection points of the edges of two 2D polygons, a simple function made to follow up a Newsgroup discussion
Curve Intersect 2 by Sebastian Hölz
This file is based on the Curve Intersect function by Duane Hanselman. It extends the scope of the function to handle arbitrary lines / polygons, which may also have vertical segments or segments with non-increasing x-values.
Curve intersections by NS
While a few other functions already exist in FEX that compute the
intersection points of curves, this short piece of code was written
with speed being the highest priority. No loops are used throughout,
taking full advantage of MATLAB's vectorization capabilities
Fast and Robust Curve Intersections by Douglas Schwarz
This function computes the (x,y) locations where two curves intersect. The curves can be broken with NaNs or have vertical segments. It is also very fast (at least on data that represents what I think is a typical application).
geom2d by David Legland
[...] derive new shapes: intersection between 2 lines, between a line and a circle, parallel and perpendicular lines
I have a convex polygon in 3D. For simplicity, let it be a square with vertices, (0,0,0),(1,1,0),(1,1,1),(0,0,1).. I need to arrange these vertices in counter clockwise order. I found a solution here. It is suggested to determine the angle at the center of the polygon and sort them. I am not clear how is that going to work. Does anyone have a solution? I need a solution which is robust and even works when the vertices get very close.
A sample MATLAB code would be much appreciated!
This is actually quite a tedious problem so instead of actually doing it I am just going to explain how I would do it. First find the equation of the plane (you only need to use 3 points for this) and then find your rotation matrix. Then find your vectors in your new rotated space. After that is all said and done find which quadrant your point is in and if n > 1 in a particular quadrant then you must find the angle of each point (theta = arctan(y/x)). Then simply sort each quadrant by their angle (arguably you can just do separation by pi instead of quadrants (sort the points into when the y-component (post-rotation) is greater than zero).
Sorry I don't have time to actually test this but give it a go and feel free to post your code and I can help debug it if you like.
Luckily you have a convex polygon, so you can use the angle trick: find a point in the interior (e.g., find the midpoint of two non-adjacent points), and draw vectors to all the vertices. Choose one vector as a base, calculate the angles to the other vectors and order them. You can calculate the angles using the dot product: A · B = A B cos θ = |A||B| cos θ.
Below are the steps I followed.
The 3D planar polygon can be rotated to 2D plane using the known formulas. Use the one under the section Rotation matrix from axis and angle.
Then as indicated by #Glenn, an internal points needs to be calculated to find the angles. I take that internal point as the mean of the vertex locations.
Using the x-axis as the reference axis, the angle, on a 0 to 2pi scale, for each vertex can be calculated using atan2 function as explained here.
The non-negative angle measured counterclockwise from vector a to vector b, in the range [0,2pi], if a = [x1,y1] and b = [x2,y2], is given by:
angle = mod(atan2(y2-y1,x2-x1),2*pi);
Finally, sort the angles, [~,XI] = sort(angle);.
It's a long time since I used this, so I might be wrong, but I believe the command convhull does what you need - it returns the convex hull of a set of points (which, since you say your points are a convex set, should be the set of points themselves), arranged in counter-clockwise order.
Note that MathWorks recently delivered a new class DelaunayTri which is intended to superseded the functionality of convhull and other older computational geometry stuff. I believe it's more accurate, especially when the points get very close together. However I haven't tried it.
Hope that helps!
So here's another answer if you want to use convhull. Easily project your polygon into an axes plane by setting one coordinate zero. For example, in (0,0,0),(1,1,0),(1,1,1),(0,0,1) set y=0 to get (0,0),(1,0),(1,1),(0,1). Now your problem is 2D.
You might have to do some work to pick the right coordinate if your polygon's plane is orthogonal to some axis, if it is, pick that axis. The criterion is to make sure that your projected points don't end up on a line.